1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Car Distance Kinematics motion

  1. Dec 11, 2013 #1
    1. The problem statement, all variables and given/known data
    You're speeding at 85km/h when you notice that you're only 10m behind the car in front of you which is moving at the legal speed limit of 60km/h. You slam on your brakes and your car negatively accelerates at 4.2m/s/s. Assuming the other car continues at constant speed, will you collide? If so, at what relative speed?
    Otherwise, what will be the distance between the cars at their closest approach?

    2. Relevant equations


    3. The attempt at a solution

    t = [vf-vi]/a
    t = [0-23.61ms^-1]/-4.2ms^-2
    t = 5.619s
    at t = 5.619s, my car would have traveled 66m
    the car moving at 60kmh^-1 would have traveled 93.65m
    No collision occurs and the minimum distance between the 2 cars is (93.65m-66m)+10m
    Last edited: Dec 11, 2013
  2. jcsd
  3. Dec 11, 2013 #2
    Can you list what equations you're using from your textbook in order to solve this?
  4. Dec 11, 2013 #3
    No equations are given.

    I've made another attempt at the solutions in my original post.
  5. Dec 11, 2013 #4
    Well there are some equations that are always true under constant acceleration which should either be given to you or you should know enough calculus to derive them.

    The one important here is: [itex]x(t)=x(0)+v(0)t+\frac{1}{2}at^{2}[/itex]

    This is an equation modeling the position of any object under constant acceleration. x(0) is the position when time is 0, and same with v(0).

    If these model position for any time t, if you were to set up an equation modeling both cars positions, if you set them equal to you each other you would be saying that for some time, t, both cars occupy the same position, x.

    So if that is solvable for a time, t, then they would have crashed. If it's unsolvable, then they never come into contact.
  6. Dec 11, 2013 #5
  7. Dec 11, 2013 #6
    The above kinematics equation brings me to the same answer as what I did too.
  8. Dec 11, 2013 #7
    In your original post, you set vf=0, this would not be true. Your final velocity would be unknown at first glance. Could you show me your work using the equation I posted?
  9. Dec 11, 2013 #8
    x(5.619) = x(0) + 23.61ms^-1 (5.619) + 0.5(-4.2ms^-2)(5.619)^2 = 66.36m
    (5.619) = x(0) + 16.67ms^-1(5.619) + 0.5(0)(5.619)^2 = 93.66m

    (93.66-66.36) + 10
  10. Dec 11, 2013 #9
    Show me your derivation of the time t at which their distance is minimized. I calculated a different t.
  11. Dec 11, 2013 #10
    t = [vf - vi]/a
    t = [(0ms^-1) - 23.61ms^-1]/-4.2ms^-2
    t = 5.619s
  12. Dec 11, 2013 #11
    Ok, your vf would not be zero. This is your problem.

    In collisions, you can tap each other but both remain in motion.

    What you found was instead the time it would take for you to stop if you maintained the same rate of deceleration.

    In order to find the minimum distance, you can construct a function OF their distance, namely x2-x1. I set mine equal to some arbitrary letter, let's say d, and then minimized it.

    What do you know about minimization?
  13. Dec 11, 2013 #12

    Could you expound on "In order to find the minimum distance, you can construct a function OF their distance, namely x2-x1. I set mine equal to some arbitrary letter, let's say d, and then minimized it. "?
  14. Dec 11, 2013 #13
    Pretty much. The point where f''>0 and f'=0 is a local minimum.

    The trick is in finding the function you're minimizing, the distance between the two cars as a function of time.
  15. Dec 11, 2013 #14
    Could you provide the mathematical reasoning behind as to how I should set up the equation? I do have a more difficult time dealing with applied mathematics than pure mathematics.
  16. Dec 11, 2013 #15
    The distance between any two points is one point minus the other point, at least on a one dimensional line.

    So if those points are changing with time, the distance function also changes with time, but it always remains one point minus the other.

    Just make those points functions of time.
  17. Dec 11, 2013 #16
    xf - xi = vit + 0.5at^2
    d' = vi + at
    -23.61ms^-1 = -4.2ms^-2 t
    t = 5.6214

    If I were to differentiate d', then wouldn't the time variable be non-existent? Can I deduce that since d" = a, then, this equation is unsolvable and therefore no collision occurs?
    Last edited: Dec 11, 2013
  18. Dec 11, 2013 #17
    You have an equation for the position of both cars. If you were to subtract one from the other you could get a varying distance between the cars as a function of time. Note you're not using the initial and final velocities, you have a function of time.

    Derive THAT equation and set it to 0.
  19. Dec 12, 2013 #18
    d1 = vit + 0.5at^2
    d2 = vit + 0.5at^2

    d1 = 16.67ms^-1 t
    d2 = 23.1,s^-1 t + 0.5(-4.2ms^-2)t^2

    d2-d1 = 23.1ms^-1 t + 0.5(-4.2ms^-2)t^2 - 16.67ms^-1 t
    d = 6.43ms^-1 t - 0.5(4.2ms^-2)t^2
    t = 1.53
    Last edited: Dec 12, 2013
  20. Dec 12, 2013 #19
    Alright, so you have a function that models position that change when time changes (it's a function of time)

    This is: [itex]x(t)=x(0)+v(0)t+\frac{1}{2}at^2[/itex]

    So for each car we can construct their own, specific function of position:


    You've done this already to find if they would occupy the same point at any time t, so I don't feel like I'm doing the problem for you by outlining them now. You set them equal to each other, then noted you're not able to solve the equation, therefore no time t exists where they occupy the same point.

    Now if we wanted to find the distance between the cars at any point, we'd calculate their positions with the above equations, then find their difference, say at time t=1s:


    So we know car 1 is at 21.5 and car 2 is at 26.7. Their distance between each other is then:


    The key here is realizing that you don't need to evaluate the functions before you plug them into your typical difference between two points equation. Just leave them at some arbitrary time t, and set them equal to some function of distance between the two, say d(t):


    Then right there, you have a function of the distance between them as a function of time. You want to find when this function has a minimum value, so you apply what you know about minimizing functions and evaluate from there.
  21. Dec 12, 2013 #20
    d(t) = 2.1t^2 + -6.9t + 10
    d'(t) = 4.2t -6.9
    t = 1.6s

    d"(t) = 4.2

    apparently, it's unsolvable.
    Am I doing it right?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted